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(A) We know that the lengths of the altitudes are given by $P_1 = \frac{2\Delta}{a}$,$P_2 = \frac{2\Delta}{b}$,and $P_3 = \frac{2\Delta}{c}$,where $\D

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$\frac{1}{R}$

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(A) We know that the lengths of the altitudes are given by $P_1 = \frac{2\Delta}{a}$,$P_2 = \frac{2\Delta}{b}$,and $P_3 = \frac{2\Delta}{c}$,where $\Delta$ is the area of the triangle.Substituting these into the expression:$\frac{\cos A}{P_1} + \frac{\cos B}{P_2} + \frac{\cos C}{P_3} = \frac{a \cos A}{2\Delta} + \frac{b \cos B}{2\Delta} + \frac{c \cos C}{2\Delta}$Using the sine rule $a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$:$= \frac{2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C}{2\Delta}$$= \frac{R(\sin 2A + \sin 2B + \sin 2C)}{2\Delta}$Using the identity $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$:$= \frac{R(4 \sin A \sin B \sin C)}{2\Delta}$Since $\Delta = \frac{abc}{4R}$,we have $\frac{1}{2\Delta} = \frac{2R}{abc}$:$= \frac{4R \sin A \sin B \sin C}{2 \cdot \frac{abc}{4R}} = \frac{8R^2 \sin A \sin B \sin C}{abc}$Substituting $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,$\sin C = \frac{c}{2R}$:$= \frac{8R^2 (\frac{a}{2R}) (\frac{b}{2R}) (\frac{c}{2R})}{abc} = \frac{8R^2 \cdot \frac{abc}{8R^3}}{abc} = \frac{1}{R}$

If $P_1, P_2$ and $P_3$ are the lengths of the altitudes drawn from the vertices $A, B$ and $C$ of $\triangle ABC$ respectively,then $\frac{\cos A}{P_1} + \frac{\cos B}{P_2} + \frac{\cos C}{P_3} =$

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